Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 41

$$\nabla f\left( {4,2} \right) = 40{\bf{i}} + 32{\bf{j}}$$

$$\eqalign{ & f\left( {x,y} \right) = 5{x^2} + {y^4};\,\,\,\,\,\left( {4,2} \right) \cr & \cr & {\text{calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^2} + {y^4}} \right] \cr & {\text{treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = 10x + 0 \cr & {f_x}\left( {x,y} \right) = 10x \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^2} + {y^4}} \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = 0 + 4{y^3} \cr & {f_x}\left( {x,y} \right) = 4{y^3} \cr & \cr & {\text{The gradient of the function }}f\left( {x,y} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr & \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr & {\text{substituting the partial derivatives, we obtain}} \cr & \nabla f\left( {x,y} \right) = 10x{\bf{i}} + 4{y^3}{\bf{j}} \cr & \cr & {\text{the gradient of }}f{\text{ at }}\left( {4,2} \right){\text{ is}} \cr & \nabla f\left( {4,2} \right) = 10\left( 4 \right){\bf{i}} + 4{\left( 2 \right)^3}{\bf{j}} \cr & \nabla f\left( {4,2} \right) = 40{\bf{i}} + 32{\bf{j}} \cr} $$